Junior Macedonian Mathematical Olympiad

From $m^3 + m^2 + 7 = (m^2 - m + 1)(m+2) + (m+5)$ it follows that $m^2 - m + 1$ is a divisor of $m+5$. Obviously, one solution is $m = -5$.

Let $m \neq -5$. Then $|m^2 - m + 1| \le |m+5|$. From $m^2 - m + 1 = (m - \frac{1}{2})^2 + \frac{3}{4} > 0$ it follows that $m^2 - m + 1 \le m + 5$ or $m^2 - m + 1 \le -m - 5$. The case $m^2 - m + 1 \le -m - 5$ is not possible ($m^2 + 6 \le 0$). From $m^2 - m + 1 \le m + 5$ it follows that $m^2 - 2m - 4 \le 0$ i.e. $(m-1)^2 \le 5$. The last inequality is satisfied for $m \in \{-1, 0, 1, 2, 3\}$. Checking each value, we conclude that $m=0$ and $m=1$ satisfy the initial condition.

Finally $m \in \{-5, 0, 1\}$.